Is ${685010}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {685010}= &&{6}\cdot100000+ \\&&{8}\cdot10000+ \\&&{5}\cdot1000+ \\&&{0}\cdot100+ \\&&{1}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {685010}= &&{6}(99999+1)+ \\&&{8}(9999+1)+ \\&&{5}(999+1)+ \\&&{0}(99+1)+ \\&&{1}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {685010}= &&\gray{6\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {6}+{8}+{5}+{0}+{1}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${685010}$ is divisible by $3$ if ${ 6}+{8}+{5}+{0}+{1}+{0}$ is divisible by $3$ Add the digits of ${685010}$ $ {6}+{8}+{5}+{0}+{1}+{0} = {20} $ If ${20}$ is divisible by $3$ , then ${685010}$ must also be divisible by $3$ ${20}$ is not divisible by $3$, therefore ${685010}$ must not be divisible by $3$.